Finding the Potential Energy Function

In summary: In b and c, it is not.In summary, the conversation discusses the concept of conservative forces and potential energy. The conversation includes discussions on how to determine if a force is conservative, how to find the corresponding potential energy, and how to verify the relationship between force and potential energy using integration. The conversation also addresses the use of integration to find potential energy along a chosen path.
  • #1
embphysics
67
0

Homework Statement


Which of the following forces is conservative? (a) F = k(x,2y,3z), (b) F=k(y,x,0), and
(c) F=k(-y,x,0). For those which are conservative, find the corresponding potential energy U, and verify by direct substitution that [itex]\vec{F} = - \nabla U[/itex]

Homework Equations


The Attempt at a Solution


I have already ascertained which force fields are conservative, and now I am trying to find the potential energy associated with those force fields.

[itex]\vec{F} = - \nabla U[/itex] implies that [itex]F_x \hat{i} + F_y \hat {j} + F_z \hat{k} = \frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial x} \hat{j} + \frac{\partial U}{\partial x} \hat{k}[/itex].

Using this fact, I set the components equal to equal other, and integrated each individual component. When I went to see if this was the correct answer, I found the answer was a linear combination of the three individual components. Why is that?
 
Physics news on Phys.org
  • #2
What did you get for U in case of (a)? Remember, U is a scalar function. It does not have components. ehild
 
  • #3
\hat{}No, I understand that U is a scalar function, that it only has a scalar value associated with every point in space (only if that particular point is in the domain of the function U, of course). For part (a), I get [itex]U= - \frac{1}{2}kx^2 + U_0[/itex]. However, the actual answer is [itex]U=-\frac{1}{2}k(x^2 +2y^2 + 3z^2)[/itex]. Like mentioned in my first post, it appears that they took the individual vector components, integrated each one, and then added them together.

Here is another way I thought of solving this problem:

[itex]W =- \Delta U = \int _{\vec{r}_0}^\vec{r} F(\vec{r}')\cdot d \vec{r}'[/itex]. I can define the potential energy to be zero at [itex]\vec{r}_0 = \langle 0,0,0 \rangle[/itex], thus measuring changes in PE relative to the origin. [itex]\vec{r} \langle x',y',z' \rangle[/itex] will be some later position of the object whom the work is done by the force. Since the force is conservative, we need not concern ourselves with the path taken to get from r_0 to r; instead, we can just integrate along the straight line connecting the two points.

[itex]\Delta U = - \int_{\vec{r}_0}^\vec{r} \langle F_x, F_y , F_z \rangle \cdot \langle dx , dy , dz \rangle = -\int_{\vec{r}_0}^\vec{r} (F_x dx + F_y dy + F_z dz)[/itex](1).

The part am I having is with the limits of integration. If the integral could somehow become [itex]\Delta U = - \int_0^x F_x dx' - \int_0^y F_y dy' - \int_0^z F_z dz'[/itex] (2), then I would get the correct answer. The problem I am having difficulty with is going from (1) to (2). How do you justify such a step?
 
Last edited:
  • #4
I don't intend to be a bother, but could someone possibly help me?
 
  • #5
embphysics said:
\

Here is another way I thought of solving this problem:

[itex]W =- \Delta U = \int _{\vec{r}_0}^\vec{r} F(\vec{r}')\cdot d \vec{r}'[/itex]. I can define the potential energy to be zero at [itex]\vec{r}_0 = \langle 0,0,0 \rangle[/itex], thus measuring changes in PE relative to the origin. [itex]\vec{r} \langle x',y',z' \rangle[/itex] will be some later position of the object whom the work is done by the force. Since the force is conservative, we need not concern ourselves with the path taken to get from r_0 to r; instead, we can just integrate along the straight line connecting the two points.

[itex]\Delta U = - \int_{\vec{r}_0}^\vec{r} \langle F_x, F_y , F_z \rangle \cdot \langle dx , dy , dz \rangle = -\int_{\vec{r}_0}^\vec{r} (F_x dx + F_y dy + F_z dz)[/itex](1).

The part am I having is with the limits of integration. If the integral could somehow become [itex]\Delta U = - \int_0^x F_x dx' - \int_0^y F_y dy' - \int_0^z F_z dz'[/itex] (2), then I would get the correct answer. The problem I am having difficulty with is going from (1) to (2). How do you justify such a step?

As U is a potential function, the integration path is arbitrary. You can reach from 0 to (x,y,z) as going first from (0,0,0) to (x,0 0) then from (x,0.0) to (x,y,0) and then from (x,y,0) to (x,y,z).

ehild
 
  • #6
You have that [itex]\dfrac{\partial u}{\partial x}= x[/itex], [itex]\dfrac{\partial u}{\partial y}= 2y[/itex], and [itex]\dfrac{\partial u}{\partial z}= 3z[/itex].

Integrating the first equation, with respect to x, we have [itex]u= x^2/2+ f(y, z)[/itex]. (Since the partial derivative is with respect to the single variable, x, and we are treating y and z as constants, the "constant of integration" may be a function of y and z.)

Since [itex]u= x^2/2+ f(y,z)[/itex], [itex]\partial u/\partial y= \partial f/\partial y= 2y[/itex]. Integrating with respect to y, [itex]f= y^2+ g(z)[/itex]. (We know that f is a function of y and z only. Since the derivative is with respect to the single variable, y, and we are treating z as a constant, the "constant of integration" may be a function of z.)

Since [itex]f(x,y)= y^2+ g(z)[/itex], [itex]u(x,y,z)= x^2/2+ f(y,z)= x^2/2+ y^2+ g(z)[/itex] and so [itex]\partial u/\partial z= dg/dz= 3z[/itex]. (Notice that since g is a function of the single variable, z, this is an ordinary derivative.) Integrating [itex]g(z)= 3z^2/2+ C[/itex]. (Since g was a function of z only, this "constant of integration really is a constant!)

Since [itex]u(x, y, z)= x^2/2+ y^2+ g(z)[/itex] we now have [itex]u(x, y, z)= x^2/2+ y^2+ 3z^2/2+ C[/itex].
 
  • #7
HallsofIvy, thank you. That was precisely what I was looking for. Although, I have one question. In post number three (one of my posts), is going from integral (1) to integral (2) incorrect? If so, why? If is it correct, how can one justify the move?
 
  • #8
embphysics said:
Although, I have one question. In post number three (one of my posts), is going from integral (1) to integral (2) incorrect? If so, why? If is it correct, how can one justify the move?

Your formula does not specify the variables other than the integration variable. How would you apply it for case b? You need to integral F along the chosen path. If it is (0,0,0)-->(x,0,0)-->(x,y,0)-->(x,y,z)
[tex]\Delta U = - \int_{0,0,0}^{x,0,0} F_x (x',0,0) dx' - \int_{x,0,0}^{x,y,0} F_y(x,y',0) dy' - \int_{x,y,0}^{x,y,z´} F_z (x,y,z') dz'[/tex]

In case a, Fx=Fx(x) Fy=Fy(y), Fz=F(z), all integrands contained only the integration variable, so your integral gave the correct result.

In case b, Fx=ky, Fy=kx. If you integral Fx with respect to x' from 0 to x, you need the value of y along the integration path. What is it?

For the path (0,0,0)-->(x,0,0)-->(x,y,0)-->(x,y,z), as before,

[tex]\Delta U = - \int_{0,0,0}^{x,0,0}( 0) dx' - \int_{x,0,0}^{x,y,0} (kx) dy' - \int_{x,y,0}^{x,y,z´}(0) dz'=-kxy +C[/tex].
ehild
 

1. What is potential energy?

Potential energy is the energy possessed by an object due to its position or configuration in a system. It is a form of stored energy that can be converted into other forms, such as kinetic energy, to do work.

2. Why is finding the potential energy function important?

Finding the potential energy function allows us to accurately describe and predict the behavior of a system. It provides a mathematical representation of the forces at play and can help us understand the stability and equilibrium of a system.

3. How do you find the potential energy function?

The potential energy function can be found by taking the negative gradient of the corresponding force function. This means finding the partial derivative of the force function with respect to each variable and then adding them together with a negative sign.

4. Can potential energy be negative?

Yes, potential energy can be negative. This typically occurs when the reference point is chosen to be at a higher potential energy level than the actual position of the object.

5. What are some common examples of potential energy?

Some common examples of potential energy include gravitational potential energy, elastic potential energy, and chemical potential energy. Gravitational potential energy is stored in objects at a height, elastic potential energy is stored in stretched or compressed springs, and chemical potential energy is stored in the bonds between atoms in molecules.

Similar threads

Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
2
Replies
44
Views
3K
  • Advanced Physics Homework Help
Replies
9
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
814
  • Advanced Physics Homework Help
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
888
  • Advanced Physics Homework Help
Replies
9
Views
1K
  • Advanced Physics Homework Help
Replies
11
Views
1K
Back
Top